Ethylene (C2H4) can be prepared by heating hexane (C6H14) according to th balanced equation. If the yield of ethylene production is 42.5%, how many moles of hexane (C6H14) must be used to produc 481g of ethylene?

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Ethylene (C2H4) can be prepared by heating hexane (C6H14) according to th balanced equation. If the yield of ethylene production is 42.5%, how many moles of hexane (C6H14) must be used to produc 481g of ethylene?

Chemistry
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The molar mass of C2H4 is 28.05 g/mol
C6H14 -> 3C2H4 + H2. 481 grams of C2H4=481/28.05 moles =17.15. Now 1 mole of hexane would give 0.425*3 moles of ethane as effeciency is 42.5%. 1 mole gives 1.725 moles of ethene. you need 17.15 moles so you would require 17.15/1.725 moles of hexane which is approximately 10 moles.

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